question_answer 29)

(a) Complete the table and by inspection of the table, find the solution to the equation \[m+10=16\].

\[m\]

1

2

3

4

5

6

7

8

9

10

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\[m+10\]

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(b) Complete the table and by inspection of the table, find the solution to the equation \[5t=35\].

\[t\]

3

4

5

6

7

8

9

10

11

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\[5t\]

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(c) Complete the table and find the solution of the equation z/3 = 4 using the table.

\[z\]

8

9

10

11

12

13

14

15

16

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\[\frac{z}{3}\]

\[2\frac{2}{3}\]

3

\[3\frac{1}{3}\]

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(d) Complete the table and find the solution to the equation \[m-7=3\].

\[m\]	5	6	7	8	9	10	11	12	13	?	?
\[m-7\]	?	?	?	?	?	?	?	?	?	?	?

Answer:

                (a) The complete table is shown below

\[m\]	\[m+10\]
1	\[1+10=11\]
2	\[2+10=12\]
3	\[3+10=13\]
4	\[4+10=14\]
5	\[5+10=15\]
6	\[6+10=16\]
7	\[7+10=17\]
8	\[8+10=18\]
9	\[9+10=19\]
10	\[10+10=20\]
11	\[11+10=21\]
12	\[12+10=22\]
13	\[13+10=23\]

By inspection of the above table, we see that \[m=6\] satisfies the equation \[m+10\text{ }=16\]. [\[\because \] at \[m=6,\] LHS = RHS] Hence, \[m=6\] is its solution. (b) The complete table is shown below

\[t\]	\[5t\]
3	\[5\times 3=15\]
4	\[5\times 4=20\]
5	\[5\times 5=25\]
6	\[5\times 6=30\]
7	\[5\times 7=35\]
8	\[5\times 8=40\]
9	\[5\times 9=45\]
10	\[5\times 10=50\]
11	\[5\times 11=55\]
12	\[5\times 12=60\]
13	\[5\times 13=65\]
14	\[5\times 14=70\]
15	\[5\times 15=75\]
16	\[5\times 16=80\]

By inspection of the above table, we find that \[t=7\] satisfies the equation \[5t=35\].                       [\[\because \]at \[t=7,\] LHS = RHS] Hence, \[t=7\] is its solution. (c) The complete table is shown below.

\[z\]	\[\frac{z}{3}\]
8	\[\frac{8}{3}=2\frac{2}{3}\]
9	\[\frac{9}{3}=3\]
10	\[\frac{10}{3}=3\frac{1}{3}\]
11	\[\frac{11}{3}=3\frac{2}{3}\]
12	\[\frac{12}{3}=4\]
13	\[\frac{13}{3}=4\frac{1}{3}\]
14	\[\frac{14}{3}=4\frac{2}{3}\]
15	\[\frac{15}{3}=5\]
16	\[\frac{16}{3}=5\frac{1}{3}\]
17	\[\frac{17}{3}=5\frac{2}{3}\]
18	\[\frac{18}{3}=6\]
19	\[\frac{19}{3}=6\frac{1}{3}\]
20	\[\frac{20}{3}=6\frac{2}{3}\].

By inspection of the above table, we find that \[t=12\] satisfies the equation\[\frac{z}{3}=4.\] [\[\because \] at \[t=12,\] LHS = EHS] Hence, \[z=12\] is its solution. (d) The complete table is shown below

\[m\]	\[m-7\]
5	\[5-7=-2\]
6	\[6-7=-1\]
7	\[7-7=0\]
8	\[8-7=1\]
9	\[9-7=2\]
10	\[10-7=3\]
11	\[11-7=4\]
12	\[12-7=5\]
13	\[13-7=6\]
14	\[14-7=7\]
15	\[15-7=8\]

By inspection of the above table, we find that \[m=10\] satisfies the equation \[m-7=3\]. [\[\because \]at \[m=10,\] LHS = RHS] Hence, \[m=10\] is its solution.